If the digits $7,7,3,2$, and 1 are randomly arranged from left to right, what is the probability both of the 7 digits are to the left of the 1 digit?
The answer is $1/3$ because $1 7 7$, $7 7 1$, $7 1 7$.
I thought about re-arranging the two sevens as one digit "77"1 then the three and two have two positions, and we re-arrange the "77" and 1 glued together using $3 \choose 1$ and divide by $5!$
On
Alternative approach:
Partition all of the ways of ordering each element in the following multiset: $\{7-a, 7-b, 3, 2, 1\}.$
Here, I am attaching a suffix to each of $~7-a,~7-b~$ to assist in distinguishing between these elements.
The natural way of creating the subsets is as follows:
___ 7-a ___ 7-b ___ 1 ___
___ 7-a ___ 1 ___ 7-b ___
___ 7-b ___ 7-a ___ 1 ___
___ 7-b ___ 1 ___ 7-a ___
___ 1 ___ 7-a ___ 7-b ___
___ 1 ___ 7-b ___ 7-a ___
Each of the $(3! = 6)$ subsets above represents a specific pattern in how the elements $7-a, 7-b,$ and $1$ are ordered. Each pattern has four gaps, before and after the three ordered elements.
In effect, the assertion is that each subset has exactly the same number of elements. To enumerate the number of elements in a specific subset, you have to enumerate how many ways that there are of ordering the extraneous elements (i.e. $3,2$) and assigning them to one or more of the four gaps in the subset's pattern.
By symmetrical considerations, there is no reason to believe that the enumeration will be different between any two of the six subsets. That is, the issue of which element, $7a,7b,$ or $1$, is in one of the wall-boundaries-between-the-gaps, should in no way affect the enumeration of the number of ways of ordering the extraneous elements, $2,3$, and assigning these extraneous elements to one or more of the four gaps.
Therefore, assuming that your intuition accepts this concept, you are justified in assuming that each subset has exactly the same number of elements.
This implies that it is irrelevant how large each subset is. That is, it is irrelevant how many ways that there are of ordering the extraneous elements $(2,3)$ and assigning these elements to one or more of the subset's four gaps.
This implies that the problem is reduced to counting how many of the $(3! = 6)$ subsets have a $1$ as the first wall-boundary-element.
As a further illustration of the concept involved, consider the following multiset
$$\{7-a,7-b,7-c,1-a,1-b,2,3,4,5,6,8,9\}.$$
Supppose that you were examining all of the ways of ordering all of the elements in the above multiset. Further suppose that you wanted to compute the probability that of the $5$ elements $\{7-a,7-b,7-c,1-a,1-b\}$, that the first such element among these five elements would be either $1-a$ or $1-b$.
Based on the concepts at the start of this answer, you have three alternative (easy) approaches:
Each of the $5$ elements is equally likely to be first. Therefore, the desired probability must be $\dfrac{2}{5}.$
There are $~\displaystyle \binom{5}{2}~$ ways of assigning the two elements $1-a,1-b$ to the five pertinent positions. Here, no attempt is made to distinguish whether 1-a comes before or after 1-b.
Further, there are $~\displaystyle \binom{4}{2}~$ ways of assigining the two elements $1-a,1-b$ to the four pertinent positions other than the first position.
Therefore, the desired probability is
$\displaystyle 1 - \frac{\binom{4}{2}}{\binom{5}{2}}.$
Of the $(5! = 120)$ ways of ordering the five pertinent elements, there are $2 \times 4!$ ways of ordering them so that either 1-a, or 1-b is the first element.
Therefore, the desired probability is
$~\displaystyle \frac{2 \times (4!)}{5!}.$
You are correct that the answer is $1/3$. Only the relative positions of the $1$ and the two $7$s matter.
If you insist on considering all five digits, there are $\binom{5}{2}$ ways to place the two $7$s and $3!$ ways to arrange the remaining three distinct digits in the remaining three positions. Therefore, there are $$\binom{5}{2}3!$$ elements in the sample space, not $5!$.
For the favorable cases, there are five ways to place the $2$ and four ways to place the $3$. Once those numbers are placed, there is only one way to arrange the $7$s and $1$ in the remaining three places so that both $7$s appear before the $1$. Hence, there are $$5 \cdot 4$$ favorable cases.
Consequently, the probability that both $7$s appear before the $1$ is $$\frac{5 \cdot 4}{\dbinom{5}{2}3!} = \frac{1}{3}$$ as you found by considering the relative positions of the two $7$s and the $1$s.